9. Roman to Integer

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本文介绍了一种将罗马数字转换为整数的算法实现，包括如何解析罗马数字字符串并根据转换规则计算出对应的整数值。

Description:

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

Conversion Rule:

基本字符	I	V	X	L	C	D	M
相应的阿拉伯数字表示为	1	5	10	50	100	500	1000

1.相同的数字连写、所表示的数等于这些数字相加得到的数、如：Ⅲ=3；
2.小的数字在大的数字的右边、所表示的数等于这些数字相加得到的数、如：Ⅷ=8、Ⅻ=12；
3.小的数字（限于 I、X 和 C）在大的数字的左边、所表示的数等于大数减小数得到的数、如：Ⅳ=4、Ⅸ=9；
4.正常使用时、连写的数字重复不得超过三次；

Solution:

1. How to split the String and express them to Integer ?

(1) String.toCharArray()

// return an array.

String.charAt(int i)

// according to the traversal, get the each Char.

(2) switch(String.char(int i)){

case 'M' :

num[I] = 1000;

break;

HashMap<Character, Integer> map = new HashMap<Character, Integer>();

2. How to calculate the result ?

If the left character's value is smaller than the right one, we should add the left one.

If the left character's value is larger than the right one or this is the last character, we should subtract the left one.

Therefore, we can use the top s.length-1 value to compare, and leave the last one to add only.

codes

import java.util.HashMap;

public class RomantoInteger {
    public static int romanToInt(String s){
        int[] nums = new int[s.length()];
        for (int i = 0; i < s.length(); i++) {
            switch (s.charAt(i)) {
                case 'M':
                    nums[i] = 1000;
                    break;
                case 'D':
                    nums[i] = 500;
                    break;
                case 'C':
                    nums[i] = 100;
                    break;
                case 'L':
                    nums[i] = 50;
                    break;
                case 'X':
                    nums[i] = 10;
                    break;
                case 'V':
                    nums[i] = 5;
                    break;
                case 'I':
                    nums[i] = 1;
                    break;
            }
        }
        int sum = 0;
        for (int i = 0; i < nums.length-1; i++){
            if (nums[i] >= nums[i+1]){
                sum += nums[i];
            }
            else {
                sum -= nums[i];
            }
        }
        return sum + nums[nums.length-1];
    }
}

Lesson learned:

1. String function

2. switch